20=-16t^2+140t+80

Solution for 20=-16t^2+140t+80 equation:

20=-16t^2+140t+80

We move all terms to the left:

20-(-16t^2+140t+80)=0

We get rid of parentheses

16t^2-140t-80+20=0

We add all the numbers together, and all the variables

16t^2-140t-60=0

a = 16; b = -140; c = -60;
Δ = b2-4ac
Δ = -1402-4·16·(-60)
Δ = 23440
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{23440}=\sqrt{16*1465}=\sqrt{16}*\sqrt{1465}=4\sqrt{1465}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-140)-4\sqrt{1465}}{2*16}=\frac{140-4\sqrt{1465}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-140)+4\sqrt{1465}}{2*16}=\frac{140+4\sqrt{1465}}{32}
$